What is the volume of the solid lying between the curves y= x^2+2x-4 and g=2x-3 ( on the left side of the y-axis) and is rotated about y-axis.
3b.jpg
What is the volume of the solid lying between the curves y= x^2+2x-4 and g=2x-3 ( on the left side of the y-axis) and is rotated about y-axis.
3b.jpg
Last edited by Subhotosh Khan; 12-25-2017 at 04:18 PM. Reason: typo
Your drawing is good.
Your integral is almost good. [tex]\pi[/tex] seems to be missing.
Now what?
"Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.
I presume the curves are supposed to be y=x^2+2x-4 and y=2x-3. But your work shows a function called f; what is that?
If you continue the way you are starting, f will have to be defined piecewise, so that your volume will be a sum of two integrals. That's doable, but messy.
This region is more easily handled by integrating with respect to x rather than y as you show. Have you learned about the "cylindrical shell" method?
There are no solids lying between two curves in a plane.
What you're trying to ask is:
What is the volume obtained by rotating, about the y-axis, the region in Quadrant III lying between the curves
f(x) = x^2 + 2x - 4
g(x) = 2x - 3
I agree with Dr. Peterson. Integrating along the x-axis is much easier, using f(x)-g(x) for the height of each shell. This approach requires one integral.
If you want to use disks and integrate along the y-axis, instead, then you'll need to determine a function for the horizontal distance between the curves in terms of y (i.e., the radius of each disk, integrating from y=-5 to y=-4). You'll then need a different function for the horizontal distance between function g and the y-axis (i.e., the radius of each disk, integrating from y=-4 to y=-3). Inverse functions for f and g will help. This approach requires summing two integrals.
"English is the most ambiguous language in the world." ~ Yours Truly, 1969
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